I would like to calculate the $\mathcal{L}_\xi(\nabla_a K^b)$ for the case where $\mathcal{L}_\xi(K^b)=0$ The only Idea that I have is that

$$\mathcal{L}_\xi(\nabla_a K^b)=\mathcal{L}_\xi(\partial_a K^b + \Gamma_{ac}^bK^c)=\mathcal{L}_\xi(\Gamma_{ac}^b) K^c $$

and using the fact that: $$\mathcal{L}_{\xi} \Gamma_{\lambda \mu}^{\rho}=\xi^\sigma\partial_\sigma \Gamma_{\lambda \mu}^{\rho}+\partial_{\lambda}\xi^{\sigma}\Gamma_{\sigma\mu}^{\rho}+\partial_{\mu}\xi^{\sigma}\Gamma_{\lambda\sigma}^{\rho}-\Gamma_{\lambda\mu}^{\sigma}\partial_{\sigma}\xi^\rho-\partial_\lambda\partial_\mu \xi^{\rho}$$

is it Ok or is there any more creative idea for that?

  • $\begingroup$ Why don't you write the Lie derivative of $\nabla_a K^b$ without expanding the covariant derivative, then use the condition you cite? $\endgroup$ – auxsvr May 21 '16 at 8:24
  • $\begingroup$ What do you mean by calculating in this case? What form of this expression do you need? What do you need it for? $\endgroup$ – Blazej May 21 '16 at 10:36
  • 1
    $\begingroup$ Hint: If your connection is torsion free as I guess, you can replace the standard derivative with the covariant one in computing the Lie derivative. Your condition ${\cal L}_\xi K=0$ can therefore be used interchanging two covariant derivatives by means of the definition of Riemann curvature tensor... $\endgroup$ – Valter Moretti May 21 '16 at 10:52

If the connection you are using is torsion-free (as the Levi-Civita one), you can systematically replace the coordinate derivative for the covariant one. So (please, pay attention to signs and positions of indexes, since I could use a convention different from yours) $${\cal L}_\xi \nabla_a K^b = \xi^c\nabla_c \nabla_a K^b - (\nabla_c\xi^b) \nabla_a K^c + (\nabla_a\xi^c) \nabla_c K^b $$ It is possible to elaborate a bit the right-hand side reducing the number of terms if passing the double derivative from $K$ to $\xi$ and exploiting your condition on $K$, now written as $$\xi^c\nabla_c K^b = K^c\nabla_c \xi^b\:.$$ Here is the procedure. $$ {\cal L}_\xi \nabla_a K^b = \xi^c\nabla_c \nabla_a K^b - (\nabla_c\xi^b) \nabla_a K^c + (\nabla_a\xi^c) \nabla_c K^b $$ $$=\xi^c\nabla_a \nabla_c K^b + \xi^c {R_{cad}}^bK^d - (\nabla_c\xi^b) \nabla_a K^c + (\nabla_a\xi^c) \nabla_c K^b$$ $$= \nabla_a(\xi^c \nabla_c K^b)- (\nabla_a\xi^c) \nabla_c K^b + \xi^c {R_{cad}}^bK^d - (\nabla_c\xi^b) \nabla_a K^c + (\nabla_a\xi^c) \nabla_c K^b$$ $$= \nabla_a(K^c\nabla_c \xi^b) - (\nabla_a\xi^c) \nabla_c K^b + \xi^c {R_{cad}}^bK^d - (\nabla_c\xi^b) \nabla_a K^c + (\nabla_a\xi^c) \nabla_c K^b$$ $$=(\nabla_a K^c)\nabla_c \xi^b + K^c\nabla_a \nabla_c \xi^b +\xi^c {R_{cad}}^bK^d - (\nabla_c\xi^b) \nabla_a K^c\:.$$ Summing up, barring mistakes and different conventions in signs and the order of indexes of Riemann tensor, $${\cal L}_\xi \nabla_a K^b = K^c\nabla_a \nabla_c \xi^b +\xi^c {R_{cad}}^bK^d $$

  • $\begingroup$ It was great. I realized that according to $\delta\Gamma_{\beta\mu}^{\alpha}=\frac{1}{2}(\nabla_{\beta}\nabla_{\mu}+\nabla_{\mu}\nabla_{\beta)}\xi^{\alpha}+\frac{1}{2}\xi^{\sigma}(R^{\alpha}_{\beta\sigma \mu}+R^{\alpha}_{\mu\sigma\beta}))$ both method have the same answer. I have another question: My final goal is to calculating $\mathcal{L}_{\psi}\mathcal{L}_{\xi}(\sqrt{-g}\nabla^{c}K^b)$. Based on our result I need to calculate something similar to $\mathcal{L}_{\psi}(\xi^c R_{cad}^{b}K^a)$ ; I have no idea how to do that, I wonder if you could possibly help me with this too! $\endgroup$ – Uncle feynman May 23 '16 at 1:14
  • $\begingroup$ Sorry I am too busy, I cannot. $\endgroup$ – Valter Moretti May 24 '16 at 13:09

It might be useful to know the formula for Lie derivative of metric itself. It is $\mathcal{L}_Xg_{ab}=\nabla_a X_b+\nabla_b X_a$. Then you might notice that covariant derivative can be split into symmetric and atisymmetric part: $$\nabla _a K_b = \nabla _{(a} K_{b)}+\nabla _{[a} K_{b]}$$ The first part is half of lie derivative of metric with respect to your field, so it contains the information about how metric changes as you flow along the integral lines of $K$. The antisymmetrized derivative is the so called exterior derivative (actually one half of exterior derivative) and in this term Christoffel symbols exactly cancel (because they are symmetric, you can easily check that). Therefore you get $$\nabla _a K_b = \frac{1}{2} \mathcal L_K g_{ab}+\partial _{[a} K_{b]}=\frac{1}{2} \left( \mathrm d K_{ab} \right)$$ where in second equality I just substituted the difnition of this exterior derivative. This $\mathrm d$ has the property that it commutes with Lie derivative. So since Lie derivative of K vanishes, Lie derivative of this term will vanish. $$\mathcal{L}_{\xi}\nabla_a K_b=\frac{1}{2}\mathcal{L}_{\xi}\mathcal{L}_{K}g_{ab}$$ This is not quite what you need yet but I thought it might be useful. It can be transformed in plenty of ways using properties of Lie derivative. I remark that this split is very natural here. One term contains information about geometry (symmetric) while antisymmetric doesn't feel it (Christoffel symbols cancel out). Word of caution: remember that you have idicies in different places than me and Lie derivative doesn't commute with lowering and raising indices so you will get extra terms containing Lie derivatives of metric.


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